Problem: Ikta has an extraordinary feeling for undirected graphs. Ikta likes the undirected graph G and its two-point s, t pair (G, s, t) that has the greatest "beauty". The "beauty" of a pair (G, s, t) is the edge e = \\ {u, v \\} (u and v are two different points of G), and the shortest path from s to t in G. Is the number of things whose length is greater than the length of the shortest path from s to t in the undirected graph with g plus e.

Your job is to write a program that seeks its "beauty" given a pair (G, s, t).



Input

The input is given in the following format.

> N M s t
> x1 y1
> ...
> xi yi
> ...
> xM yM
>

First, the number of vertices, the number of edges, and the integers N, M, s, t representing the two vertices of the undirected graph are input. From the 2nd line to the M + 1th line, two vertices connected by edges are input. (However, let the set of G vertices be \\ {1, ..., N \\}.)

Constraints

Each variable being input satisfies the following constraints.

* 2 ≤ N ≤ 100,000
* 1 ≤ M ≤ 300,000
* 1 ≤ s, t, xi, yi ≤ N
* Different from s and t
* Guaranteed to reach t from s

Output

When the given graph is G, output the "beauty" of the set (G, s, t) in one line.

Examples

Input

3 2 1 3
1 2
2 3


Output

1


Input

9 8 7 8
2 6
4 9
8 6
9 2
3 8
1 8
8 5
7 9


Output

7


Input

4 3 1 4
1 2
3 4
4 1


Output

0


Input

9 7 8 9
9 6
6 5
3 6
3 7
2 5
8 5
1 4


Output

2

Solution (in CPP):

Solution:
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
#include <utility>
using namespace std;

typedef long long LL;
typedef pair<int,int> pii;

int bfs(int s, int t, const vector<vector<int> > &G, vector<int> &cnt){
	int n = G.size();
	vector<int> ds(n, -1);
	ds[s] = 0;
	queue<int> q;
	q.push(s);
	q.push(-1);
	int tm = 1;
	while(q.size() > 1){
		int u = q.front();
		q.pop();
		if(u == -1){
			q.push(-1);
			++tm;
		}
		else{
			for(int v : G[u]){
				if(ds[v] == -1){
					ds[v] = tm;
					q.push(v);
				}
			}
		}
	}

	cnt.assign(n, 0);
	for(int i = 0; i < n; ++i){
		if(ds[i] != -1){ ++cnt[ds[i]]; }
	}

	return ds[t];
}

int main(){
	int n, m, s, t;
	scanf("%d%d%d%d", &n, &m, &s, &t);
	vector<vector<int> > G(n + 1), rG, tmp;
	vector<int> cnt1, cnt2;
	for(int i = 0; i < m; ++i){
		int x, y;
		scanf("%d%d", &x, &y);
		G[x].push_back(y);
		G[y].push_back(x);
	}

	int ds = bfs(s, t, G, cnt1);
	bfs(t, s, G, cnt2);
	LL ans = 0;
	for(int i = 0; i < n; ++i){
		if(ds - i - 2 >= 0){
			ans += (LL)cnt1[i] * cnt2[ds - i - 2];
		}
	}
	printf("%lld\n", ans);
}